Answer :
Let's solve the equation [tex]\(\sqrt{18x + 5} = 7\)[/tex] step-by-step.
### Step 1: Isolate the Square Root
The given equation is:
[tex]\[ \sqrt{18x + 5} = 7 \][/tex]
### Step 2: Square Both Sides to Eliminate the Square Root
To remove the square root, square both sides of the equation:
[tex]\[ (\sqrt{18x + 5})^2 = 7^2 \][/tex]
This simplifies to:
[tex]\[ 18x + 5 = 49 \][/tex]
### Step 3: Solve for [tex]\(x\)[/tex]
Now, solve the resulting linear equation for [tex]\(x\)[/tex]:
1. First, isolate the term containing [tex]\(x\)[/tex] by subtracting 5 from both sides:
[tex]\[ 18x + 5 - 5 = 49 - 5 \][/tex]
Simplifying this, we get:
[tex]\[ 18x = 44 \][/tex]
2. Next, solve for [tex]\(x\)[/tex] by dividing both sides by 18:
[tex]\[ x = \frac{44}{18} \][/tex]
3. Simplify the fraction:
[tex]\[ x = \frac{22}{9} \][/tex]
Thus, the candidate solution is:
[tex]\[ x = \frac{22}{9} \][/tex]
### Step 4: Check for Extraneous Solutions
To determine if this solution is extraneous, substitute [tex]\(x = \frac{22}{9}\)[/tex] back into the original equation [tex]\(\sqrt{18x + 5} = 7\)[/tex] and verify if both sides are equal.
Substitute [tex]\(x = \frac{22}{9}\)[/tex] into the left-hand side:
[tex]\[ \sqrt{18 \left(\frac{22}{9}\right) + 5} \][/tex]
Calculate inside the square root:
[tex]\[ 18 \left(\frac{22}{9}\right) = 2 \times 22 = 44 \][/tex]
So, the left-hand side becomes:
[tex]\[ \sqrt{44 + 5} = \sqrt{49} = 7 \][/tex]
Since both sides of the original equation are equal, [tex]\(x = \frac{22}{9}\)[/tex] is not an extraneous solution.
### Conclusion
The solution to the equation [tex]\(\sqrt{18x + 5} = 7\)[/tex] is:
[tex]\[ x = \frac{22}{9} \][/tex]
This solution is valid and not extraneous, as verified by substitution back into the original equation.
### Step 1: Isolate the Square Root
The given equation is:
[tex]\[ \sqrt{18x + 5} = 7 \][/tex]
### Step 2: Square Both Sides to Eliminate the Square Root
To remove the square root, square both sides of the equation:
[tex]\[ (\sqrt{18x + 5})^2 = 7^2 \][/tex]
This simplifies to:
[tex]\[ 18x + 5 = 49 \][/tex]
### Step 3: Solve for [tex]\(x\)[/tex]
Now, solve the resulting linear equation for [tex]\(x\)[/tex]:
1. First, isolate the term containing [tex]\(x\)[/tex] by subtracting 5 from both sides:
[tex]\[ 18x + 5 - 5 = 49 - 5 \][/tex]
Simplifying this, we get:
[tex]\[ 18x = 44 \][/tex]
2. Next, solve for [tex]\(x\)[/tex] by dividing both sides by 18:
[tex]\[ x = \frac{44}{18} \][/tex]
3. Simplify the fraction:
[tex]\[ x = \frac{22}{9} \][/tex]
Thus, the candidate solution is:
[tex]\[ x = \frac{22}{9} \][/tex]
### Step 4: Check for Extraneous Solutions
To determine if this solution is extraneous, substitute [tex]\(x = \frac{22}{9}\)[/tex] back into the original equation [tex]\(\sqrt{18x + 5} = 7\)[/tex] and verify if both sides are equal.
Substitute [tex]\(x = \frac{22}{9}\)[/tex] into the left-hand side:
[tex]\[ \sqrt{18 \left(\frac{22}{9}\right) + 5} \][/tex]
Calculate inside the square root:
[tex]\[ 18 \left(\frac{22}{9}\right) = 2 \times 22 = 44 \][/tex]
So, the left-hand side becomes:
[tex]\[ \sqrt{44 + 5} = \sqrt{49} = 7 \][/tex]
Since both sides of the original equation are equal, [tex]\(x = \frac{22}{9}\)[/tex] is not an extraneous solution.
### Conclusion
The solution to the equation [tex]\(\sqrt{18x + 5} = 7\)[/tex] is:
[tex]\[ x = \frac{22}{9} \][/tex]
This solution is valid and not extraneous, as verified by substitution back into the original equation.